The sides of a rhombus are parallel to the lines `x+y-1=0` and `7x-y-5=0.` It is given that the diagonals of the rhombus intersect at (1, 3) and one vertex, `A` of the rhombus lies on the line `y=2x` . Then the coordinates of vertex `A` are `(8/5,(16)/5)` (b) `(7/(15),(14)/(15))` `(6/5,(12)/5)` (d) `(4/(15),8/(15))`

Options

A. (8/5, 16/5)
B. (7/15, 14/15)
C. (6/5,12/5)
D. (4/15, 8/15)

Answer

A, C

Explanation

It is clear that the diagonals of the rhombus will be parallel to the bisectors of the given lines and will pass through (1,3). The equations of bisectors of the given lines are
`(x+y-1)/(sqrt(2)) = +-((7x-y-5)/(5sqrt(2)))`
or 2x-6y= 0, 6x+2y=5
Therefore, the equations of diagonals are x-3y+8 = 0 and 3x+y-6=0. Thus, the required vertex will be the point where these lines meet the line y=2x. Solving these lines, we get the possible coordinates as (8/5, 16/5) and (6/5, 12/5).